Chemical Kinetics Chapter-Wise Test 15

Correct answer Carries: 4.

Wrong Answer Carries: -1.

The rate constant of a reaction is \( 5.0 \times 10^{-4} \, \text{s}^{-1} \) at 350 K and \( 2.0 \times 10^{-3} \, \text{s}^{-1} \) at 360 K. What is the activation energy (\( R = 8.314 \, \text{J mol}^{-1} \text{K}^{-1} \))?

Using \( \log \frac{k_2}{k_1} = \frac{E_a}{2.303 R} \left( \frac{T_2 - T_1}{T_1 T_2} \right) \).

\( \log \frac{2.0 \times 10^{-3}}{5.0 \times 10^{-4}} = \frac{E_a}{2.303 \times 8.314} \left( \frac{10}{350 \times 360} \right) \).

\( \log 4 = 0.602 \), \( E_a = \frac{0.602 \times 19.147 \times 126000}{10} \approx 145,200 \, \text{J mol}^{-1} \approx 145.2 \, \text{kJ mol}^{-1} \).

72.6 kJ mol\(^{-1}\)
96.8 kJ mol\(^{-1}\)
145.2 kJ mol\(^{-1}\)
193.6 kJ mol\(^{-1}\)
3

A reaction’s rate constant increases from \( 2.5 \times 10^{-4} \, \text{s}^{-1} \) at 27°C to \( 7.5 \times 10^{-4} \, \text{s}^{-1} \) at 37°C. What is the activation energy (\( R = 8.314 \, \text{J mol}^{-1} \text{K}^{-1} \))?

\( T_1 = 300 \, \text{K} \), \( T_2 = 310 \, \text{K} \).

\( \log \frac{7.5 \times 10^{-4}}{2.5 \times 10^{-4}} = \frac{E_a}{2.303 \times 8.314} \left( \frac{10}{300 \times 310} \right) \).

\( \log 3 = 0.477 \), \( E_a = \frac{0.477 \times 19.147 \times 93000}{10} \approx 84,900 \, \text{J mol}^{-1} \approx 84.9 \, \text{kJ mol}^{-1} \).

42.5 kJ mol\(^{-1}\)
63.7 kJ mol\(^{-1}\)
84.9 kJ mol\(^{-1}\)
106.1 kJ mol\(^{-1}\)
3

A zero-order reaction has a rate constant of \( 5.0 \times 10^{-3} \, \text{mol L}^{-1} \text{min}^{-1} \). If the initial concentration is \( 0.25 \, \text{mol L}^{-1} \), how long will it take for 60% decomposition?

For zero-order, \( k = \frac{[\text{R}]_0 - [\text{R}]}{t} \).

60% decomposition means \( [\text{R}] = 0.4 \times 0.25 = 0.1 \, \text{mol L}^{-1} \).

\( t = \frac{0.25 - 0.1}{5.0 \times 10^{-3}} = \frac{0.15}{5.0 \times 10^{-3}} = 30 \, \text{min} \).

30 min
40 min
50 min
60 min
1

A reaction’s rate constant doubles when the temperature increases from 350 K to 360 K. What is the activation energy (\( R = 8.314 \, \text{J mol}^{-1} \text{K}^{-1} \))?

\( \log \frac{k_2}{k_1} = \frac{E_a}{2.303 R} \left( \frac{T_2 - T_1}{T_1 T_2} \right) \).

\( \log 2 = \frac{E_a}{2.303 \times 8.314} \left( \frac{10}{350 \times 360} \right) \).

\( 0.301 = \frac{E_a \times 10}{19.147 \times 126000} \), \( E_a \approx 72,600 \, \text{J mol}^{-1} \approx 72.6 \, \text{kJ mol}^{-1} \).

36.3 kJ mol\(^{-1}\)
54.5 kJ mol\(^{-1}\)
90.8 kJ mol\(^{-1}\)
72.6 kJ mol\(^{-1}\)
4

A reaction has the rate law \( \text{Rate} = k[\text{A}][\text{B}]^2 \). If the concentration of A is tripled and B is halved, what happens to the rate?

Initial rate = \( k [\text{A}] [\text{B}]^2 \).

New rate = \( k (3[\text{A}]) \left( \frac{[\text{B}]}{2} \right)^2 = k \times 3 [\text{A}] \times \frac{[\text{B}]^2}{4} = \frac{3}{4} k [\text{A}] [\text{B}]^2 \).

Rate becomes \( \frac{3}{4} \) times the initial rate.

Increases by 3 times
Decreases to \( \frac{3}{4} \) times
Remains same
Increases by \( \frac{3}{2} \) times
2

For the reaction \( 3\text{A} + \text{B} \to 2\text{C} \), the rate of disappearance of A is \( 0.12 \, \text{mol L}^{-1} \text{s}^{-1} \). What is the rate of formation of C?

Rate = \( -\frac{1}{3} \frac{\Delta[\text{A}]}{\Delta t} = \frac{1}{2} \frac{\Delta[\text{C}]}{\Delta t} \).

Given: \( -\frac{\Delta[\text{A}]}{\Delta t} = 0.12 \, \text{mol L}^{-1} \text{s}^{-1} \).

Rate = \( \frac{0.12}{3} = 0.04 \), \( \frac{\Delta[\text{C}]}{\Delta t} = 2 \times 0.04 = 0.08 \, \text{mol L}^{-1} \text{s}^{-1} \).

0.04 mol L\(^{-1}\) s\(^{-1}\)
0.08 mol L\(^{-1}\) s\(^{-1}\)
0.12 mol L\(^{-1}\) s\(^{-1}\)
0.16 mol L\(^{-1}\) s\(^{-1}\)
1

A first-order gaseous reaction has an initial pressure of 0.8 atm. After 30 s, the total pressure is 1.0 atm. What is the rate constant?

For \( \text{A} \to \text{B} + \text{C} \), \( p_t = p_i + x \), \( x = 1.0 - 0.8 = 0.2 \, \text{atm} \).

\( p_A = p_i - x = 0.8 - 0.2 = 0.6 \, \text{atm} \).

\( k = \frac{2.303}{30} \log \frac{0.8}{0.6} = \frac{2.303 \times 0.125}{30} \approx 0.0096 \, \text{s}^{-1} \).

0.0077 s\(^{-1}\)
0.0115 s\(^{-1}\)
0.0096 s\(^{-1}\)
0.0134 s\(^{-1}\)
3

A reaction’s rate increases by 2 times when the temperature rises from 25°C to 35°C. What is the activation energy (\( R = 8.314 \, \text{J mol}^{-1} \text{K}^{-1} \))?

\( T_1 = 298 \, \text{K} \), \( T_2 = 308 \, \text{K} \).

\( \log 2 = \frac{E_a}{2.303 \times 8.314} \left( \frac{10}{298 \times 308} \right) \).

\( 0.301 = \frac{E_a \times 10}{19.147 \times 91684} \), \( E_a \approx 52,800 \, \text{J mol}^{-1} \approx 52.8 \, \text{kJ mol}^{-1} \).

26.4 kJ mol\(^{-1}\)
39.6 kJ mol\(^{-1}\)
52.8 kJ mol\(^{-1}\)
66.0 kJ mol\(^{-1}\)
3

A zero-order reaction reduces the concentration from \( 0.18 \, \text{mol L}^{-1} \) to \( 0.09 \, \text{mol L}^{-1} \) in 45 s. What is the rate constant?

For zero-order, \( k = \frac{[\text{R}]_0 - [\text{R}]}{t} \).

Given: \( [\text{R}]_0 = 0.18 \, \text{mol L}^{-1} \), \( [\text{R}] = 0.09 \, \text{mol L}^{-1} \), \( t = 45 \, \text{s} \).

\( k = \frac{0.18 - 0.09}{45} = \frac{0.09}{45} = 2.0 \times 10^{-3} \, \text{mol L}^{-1} \text{s}^{-1} \).

1.0 \(\times\) 10\(^{-3}\) mol L\(^{-1}\) s\(^{-1}\)
1.5 \(\times\) 10\(^{-3}\) mol L\(^{-1}\) s\(^{-1}\)
2.0 \(\times\) 10\(^{-3}\) mol L\(^{-1}\) s\(^{-1}\)
2.5 \(\times\) 10\(^{-3}\) mol L\(^{-1}\) s\(^{-1}\)
3

For the reaction \( \text{A} + 2\text{B} \to \text{C} + \text{D} \), the rate of formation of D is \( 0.04 \, \text{mol L}^{-1} \text{s}^{-1} \). What is the rate of disappearance of B?

Rate = \( -\frac{1}{2} \frac{\Delta[\text{B}]}{\Delta t} = \frac{\Delta[\text{D}]}{\Delta t} \).

Given: \( \frac{\Delta[\text{D}]}{\Delta t} = 0.04 \, \text{mol L}^{-1} \text{s}^{-1} \).

\( -\frac{\Delta[\text{B}]}{\Delta t} = 2 \times 0.04 = 0.08 \, \text{mol L}^{-1} \text{s}^{-1} \).

0.02 mol L\(^{-1}\) s\(^{-1}\)
0.08 mol L\(^{-1}\) s\(^{-1}\)
0.04 mol L\(^{-1}\) s\(^{-1}\)
0.16 mol L\(^{-1}\) s\(^{-1}\)
2

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